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I need help with these practice problems. If anyone knows how to do them, I would be very appreciative if they would help.
{Xn} is a sequence.
1) Suppose that {Xn} converges to L and that L > 0. Prove that there exists a positive number E and a positive integer N such that Xn ≥ E for all n ≥ N.
2) Let {Xn} be a sequence of real numbers that converges to infinity.
(a) Prove that {Xn} is bounded below.
(b) Let {An} be another sequence. Suppose there is a positive integer N such that An ≥ Xn for all n ≥ N. Prove that {An} converges to infinity.
4) Let {An} and {Bn} be two number sequences. Suppose {An} is increasing, {Bn} is decreasing, and An≤ Bn for all n.
(a) Prove that both sequences converge.
(b) Must both sequences converge to a same limit? Give proof or disproof.
5) Let {An} be a sequence of nonnegative real numbers that converges to A. Prove that the sequence {√An} converges to √A.
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1.
Set ε (from the definition of convergence) equal to L/2, and try working from there.
2.
Something diverges to infinity if for all real M, there exists an N such that n > N ⇒ x_n > M.
Set M equal to (say) 1. Then by the above definition, an N exists such that n > N ⇒ x_n > 1.
Now consider the set {x_n| n ≤ N}. This is a finite set, and so has a minimum. If the minimum of that set is less than 1, then that is your lower bound. Otherwise, 1 is.
4a.
If a sequence is increasing and bounded above, then it is convergent. We're told that A_n is increasing, so to prove convergence we need to show that it is bounded above.
Now, since B_n is decreasing, B_n ≤ B_1, for all n. Hence, A_n ≤ B_1, for all n.
B_1 is therefore an upper bound for {A_n} and so {A_n} is convergent.
You can use a similar argument to prove the convergence of {B_n}.
4b.
The sequences don't necessarily converge to the same limit. It isn't too hard to find a counter-example.
Why did the vector cross the road?
It wanted to be normal.
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